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4 years ago

Soluble or Insoluble in water: Dirt

Chemistry

1 answer:

Varvara68 [4.7K]4 years ago

5 0

Answer:

Insoluble

Explanation:

Like other non-polar molecules such as petrol, wax and grease, most food and dirt is not soluble in water.

https://www.primaryconnections.org.au/sites/all/modules/primaryconnections/includes/SBR/data/Chem/sub/soap/soap.htm#:~:targetText=Like%20other%20non%2Dpolar%20molecules,is%20not%20soluble%20in%20water.

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Which reactant will be in excess, and how many moles of it will be left over?

timofeeve [1]

Answer:

Oxygen with 0.36 moles left over

Explanation:

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3 years ago

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A 250.0 mL sample of air at 100.0 K is warmed to 200.0 k at constant pressure .What is the volume of the air sample at the new t

hoa [83]

Answer:

500.0 mL

Explanation:

From <em>Charles’ Law</em>, we know that

$\frac{V_{1} }{T_{1} }= \frac{V_{2} }{T_{2 }}$

We can solve this equation to get

$V_{2} = V_{1} \times \frac{ T_{2}}{ T_{1}}$

$V_{2} = \text{250.0 mL} \times \frac{\text{200.0 K}}{\text{100.0 K}} = \textbf{500.0 mL}$

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3 years ago

The Process of forming a new<br> eukaryotic cell is called

gulaghasi [49]

Answer:

Mitosis I think

Explanation:

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3 years ago

How many grams of sodium formate, NaCHO2, would have to be dissolved in 3.0 L of 0.12 M formic acid (pKa 3.74) to make the solut

kolbaska11 [484]

Answer:

$Mass_{sodium\ formate}= 889.57\ g$

Explanation:

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the acidic buffer solution as:

$pH=pK_a+log\frac{[salt]}{[acid]}$

Given that:-

[Acid] = 0.12 M

Volume = 3.0 L

pKa = 3.74

pH = 5.30

So,

$5.30=3.74+log\frac{[sodium\ formate]}{0.12}$

Solving, we get that:-

[Sodium formate] = 4.36 M

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

So,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

So, Moles of sodium formate = 4.36*3.0 moles = 13.08 moles

Molar mass of sodium formate = 68.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$13.08\ mole= \frac{Mass}{68.01\ g/mol}$

$Mass_{sodium\ formate}= 889.57\ g$

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4 years ago

CH4 +202 — CO2 + 2H2O

lutik1710 [3]

Explanation:

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3 years ago